FIRST QUARTER MODULE 8
LIMITING REACTANT
Physical Science (Core Subject) – Grade 11/12
Quarter 1 – Module 8: Limiting Reactant
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Regional Director: Gilbert T. Sadsad
Assistant Regional Director: Jessie L. Amin
Development Team of the Module
Writer: Marileah R. Mendina
Editors: Jocelyn P. Navera, Kristina N. Nieves, Bebelyn B. Nocomora & Brenly B. Mendoza
Language Editor: Diana L. Desuyo
Reviewers: Jocelyn P. Navera, Kristina N. Nieves, Bebelyn B. Nocomora & Brenly B.
Mendoza
Illustrator: Ray Daniel G. Peralta
Layout Artist: Jose P. Gamas Jr.
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Determine the limiting reactant in a
reaction and calculate the amount of
product formed.
(S11/12PS-IIIh-27)
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Supplementary Learning Module for Senior High School Learners
LESSON
LIMITING REACTANT
When does a chemical reaction stop? When a
chemist carries out a reaction, the reactants are usually
not present in exact stoichiometric amounts that is in
the proportions indicated by the balanced equation. The
goal of a reaction is to produce the maximum quantity of
a useful compound from the starting materials;
frequently a large excess of one reactant is supplied to
ensure that the more expensive reactant is completely
converted to the desired product.
This whole process will explain why one reactant will be completely consumed before
the other runs out.
Can you now figure out when does a chemical reaction stop?
Learning the concepts related to chemical
reactions will be your key to unlock the world of
stoichiometry. Are you now ready to learn?
Make sure to perform all the activities diligently to
ensure full understanding of the lesson. Rest assured
this will be an exciting and fun-filled learning adventure.
At the end of this module, you should be able to:
•
•
Determine the limiting reactant in a
chemical reaction
Calculate the amount of product formed
from given chemical reactions.
Directions: Identify what is asked or described in each
item then write the letter of the correct answer.
1. In the preparation of a mini roast, there are 16 button mushrooms, 12 hotdogs, and 4
skewers are available. Each skewer should contain 3 mushrooms and 3 hotdogs.
How many mini roasts can be assembled?
A. 3
B. 4
C. 5
D. 6
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2. In question no.1, what ingredient is considered as the limiting reactant?
A. Mushroom
B. Hotdog
C. Both mushroom and hotdog
D. None of the above
3. Which of the following is true about the limiting reactant?
The limiting reagent in a chemical reaction is one that _____________________.
A. has the largest molar mass (formula weight).
B. has the smallest coefficient.
C. is consumed completely.
D. is in excess.
4. A propane grill on a backyard patio is left burning for too long and eventually goes
out. Which of the following chemicals is the limiting reagent?
A. Water vapor
B. Oxygen gas
C. Carbon Dioxide
D. Propane
5. In a balanced chemical equation, limiting reactant is identified by _______________.
A. calculating the number of atoms of each reactant.
B. looking at the number of moles of each reactant.
C. calculating the number of atoms of each product.
D. looking at the number of moles of each product.
Hi! How did you find the test?
Please check your answers at the answer key section and
see how you did. Don’t worry if you got a low score, this
just means that there are more things that you can learn
from this module. So, hop on!
BUILD Me Up!
Direction: Analyze the situation, perform the task and answer the
questions below.
Situation: Pretend you have a job building tricycles. The available materials are 100 handle
bars, 150 wheels, 250 pedals, and 75 seats.
Task: Draw or illustrate a tricycle with labelled parts such as handle bar, wheels, pedal and
seat.
Questions:
1. How many tricycles could you build?
2. What material limits the number of tricycles that you can build?
3. How many pedals are left over after you have built the tricycles?
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Good job in finishing the activity! Now, take time to process the information you
acquired by accomplishing the next task.
In stoichiometric calculations involving limiting reagents, the first step is to determine
which reactant is the limiting reagent. How can you determine which is the limiting reactant?
Consider the illustration below:
(Source: Chemistry by R. Chang.
Mass Relationships in Chemical
Reaction.ppt,2006)
Notice that the molecules of NO and O2 are the reactants involved in the chemical
reaction and NO2 is the product formed.
Writing the balanced chemical equation for this reaction, it will be
2NO + O2
2NO2
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Guide questions:
1. Which of the two reactants is no longer visible/present after the reaction is
completed? What do you call this kind of reactant?
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2. Which of the reactants is still visible/present after the reaction is completed? What do
they call this kind of reactant?
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The terms you have identified in the activity are called limiting reactant/reagent and
excess reactant/reagent. The reactant that produces fewer moles of product is the limiting
reagent because it limits the amount of product that can be formed. When this reactant is
used up, no more product can be formed. This lesson will teach you how to convert the
amount of reactant to amount of product.
A chemical reaction stops when one of the reactants is completely used up. In most
chemical experiments, the reactants are not in exact ratios specified by the balanced
chemical equation. Often, one of the reactants is used in excess, and the reaction is allowed
to proceed until one of the reactants is used up. The reactant that is completely used up in a
chemical reaction is called limiting reactant. As its name implies, it limits the extent of the
reaction, and thereby determines the amount of product that will be formed. On the other
hand, the reactant that still remains after the reaction stops is called the excess reactant.
As its name implies, it is in excess in the reaction.
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When more than one reactant quantity is given in a problem, it is likely that one of the
reactants will be consumed completely (the limiting reactant) while the other reactant is not
(the excess reactant). Stoichiometry is used in mass to mass calculations to easily
determine the limiting reagent and excess reagent in the chemical reaction.
Observe and study how mass-mass calculations are executed in the following
problems:
But wait!
Make sure you have a copy of the Periodic
table of Elements to easily determine the
atomic weight of a certain element. You will
use it in computing for the Molar Mass of a
molecule/substance.
PROBLEM 1:
Fe2O3 + 3CO → 2Fe + 3CO2
In the chemical equation provided above, let us suppose that a scientist was given
30.0 grams of Fe2O3 and 16.80 grams of CO simultaneously. Which substance should be
used completely and which substance would have an excess after reaction?
Step 1. Find the amount of the reactants needed in the reaction based from given amount
given in the problem.
Solution A. Using the given mass of Fe2O3, find the mass of CO needed in the reaction.
Mass of CO = 30.0 g Fe2O3 x 1.00 mole Fe2O3
160 g Fe2O3
x 3.00 moles CO
x
1.00 mole Fe2O3
28.0 g CO
1.00 mole CO
= 15.75 or 15.8 g CO
Solution B. Using the given mass of CO, find the mass of Fe2O3 needed in the reaction.
Mass of Fe2O3 = 16.80 g CO x 1.00 mole CO x 1.00 mole Fe2O3 x 160 g Fe2O3
28.0 g CO
3.00 moles CO
1.00 mole Fe2O3
= 32.0 g Fe2O3
Observe how solution A & B
use the general formula for
mas-mass calculation below:
Stoichiometry Checkpoint!
Mass A x
Molar mass A x
no. of moles B (Mole ratio from balance equation)
no. of moles A
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x Molar Mass B
Step 2. Analyze the computed value from the given value.
From the calculation, it can be deduced that 15.75 g CO is needed to react with 30.0
g Fe2O3 while 32.0 g Fe2O3 is needed to react with 16.80 g CO. It should be noted
that only 30.0 g Fe2O3 is available for the reaction.
Step 3. Identify the limiting reagent and the excess reagent.
The amount of Fe2O3 limits the reaction, hence Fe2O3 is the limiting reagent. It
follows that CO is the excess reagent.
Step 4. Compute the excess value of the excess reactant.
What is the excess value? Simply subtract the calculated amount from the given
amount. In this example, it is the amount in Solution A.
Excess value of CO = 16.80 g CO – 15.75 g CO = 1.05 g in excess
PROBLEM 2:
Consider the reaction below:
K2O + 2HCl → 2KCl + H2O
If 10.0 g K2O is made to react with 10.0 g HCl, how many grams of H2O will be
formed?
Use the following molar masses: K2O=94g/mol, HCl=36.5g/mol, and H2O=18g/mol.
How to Do
Step 1: Convert the mass of both reactants to their corresponding moles.
Moles K2O:
molesK2O=massK2O / molarmassK2O
molesK2O=10.0g K2O / 94g/mol
molesK2O=0.106molK2O
Moles HCl:
molesHCl=massHCl / molarmassHCl
molesHCl=10.0g HCl / 36.5g/mol
molesHCl=0.274molHCl
Step 2: Calculate the moles of product that can be formed from each reactant.
Moles H2O from K2O:
molesH2O=(0.106moles K2O) (1moleH2O / 1moleK2O)
molesH2O=0.106 molesH2O
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Moles H2O from HCl:
molesH2O=(0.274molesHCl) (1 moleH2O / 2 moles HCl)
molesH2O=0.137molesH2O
Step 3: Compare the moles of product formed from each reactant.
Moles H2O from K2O is less than moles H2O from HCl.
Step 4: Tag the reactant that gives a lower amount of product as the limiting reactant.
Since moles H2O from K2O is less than moles H2O from HCl, K2O is the limiting reactant.
Step 5: Convert the moles of product to mass formed from the limiting reactant. The
calculated mass is the maximum amount of product that you can form from the reaction.
massH2O=(molesH2O)(molarmassH2O)
massH2O=(0.106 mol H2O)(18g H2O/ 1 mol H2O)
massH2O=1.908 g H2O
PROBLEM 3:
Mass (g) Al x Molar mass Al x
no. of moles Fe2O3 (Mole ratio from balance equation)
no. of moles Al
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x Molar Mass Fe2O3
Mass (g) Al x Molar mass Al x
no. of moles Al2O3 (Mole ratio from balance equation)
no. of moles Al
x Molar Mass Al2O3
PROBLEM 4:
Consider the balanced chemical equation below,
1. What is the mole ratio of the reactants?
2. If 1 mole of N2 were consumed in the reaction, how many moles of NH3 were
produced?
3. What mass of NH3 will be produced if 0.350 g of H2 reacted completely with N2?
Answer/Solution:
1. Mole ratio of H2 and N2 is 3:1 (as indicated by the number before the chemical
formula)
2. 2 moles of NH3 (as reflected in the mole ratio of N2 and NH3)
3. Calculate the mass of NH3 using the given mass of H2,
Mass H2 x
Molar mass H2 x
no. of moles NH3 (Mole ratio from balance equation)
no. of moles H2
Mass of NH3 = 0.350 g H2 x 1.00 mole H2
x 2.00 moles NH3
2.00 g H2
3.00 moles H2
= 1.98 g NH3
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x
17.0 g NH3
1.00 mole NH3
x Molar Mass NH3
Helping Tips: Notice how the problems are solved in different ways. There is actually no
single strategy in solving the problems. It’s up to you to choose the strategy that works best
for you.
Now, let’s have the concept checkpoint!
Do you now understand how to determine the limiting reagent and calculate the
maximum amount of product that can be formed in a chemical reaction? Do you think you
can apply the steps in solving related problems? Let us find out in the succeeding activity.
Direction: Solve the problems below:
1. Consider the chemical reaction below:
Fe2O3 + 3CO → 2Fe + 3CO2
Suppose fifteen (15) grams of Fe2O3 was allowed to react with CO,
a. How much iron (Fe) will be produced from the reaction?
b. If 8.4 g of CO completely reacted with Fe2O3, how much iron (Fe) will be
produced from the reaction?
2. Consider the reaction
3. In the balanced chemical equation Si+2Cl2 → SiCl4, 56 g of silicon is combined with
35 g of chlorine gas in a reaction vessel:
a. How many moles of SiCl4 are formed from Cl2?
b. What is the limiting reactant?
c. How many moles of the excess reactant are left?
4. In the reaction
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5. When a mixture of 38 g of CS2 reacts with 40 g of CaO in the reaction
a. What is the limiting reagent and the excess reagent?
b. What is the excess mass of the excess reagent?
How did you find the activity? Make sure to
check your answers on the answer key. If you got 3
problems answered correctly, congratulations! If not,
try to go back to the problems once again.
This is now the end of the supplementary
learning module! You have just had an amazing and
exciting learning journey and for sure, you will have
the same in the succeeding modules.
This time, share to the class your takeaway
insights from this module by completing the following
sentence prompts.
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To strengthen what you have learned in this module, perform the activity below.
You are creative. Limits no more!
Task:
Create your own step-by-step strategy for the calculations involved in stoichiometry
particularly on limiting reactant and excess reagent. You may use songs, raps or chants or
any creative way that you can think of, in outlining the different steps. Use the
abovementioned sample problems as your guide.
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The following terms used in this module are defined as
follows:
Excess Reagent – the reactant present in quantities greater than necessary to react with
the quantity of the limiting reagent.
Limiting Reagent – the reactant used up first in a reaction and determines the amount of
product formed.
Mass-Mass Calculation – an unknown mass of one substance can be calculated from the
known mass of another substance using the general formula:
Mass A x
Molar mass A x
no. of moles B (Mole ratio from balance equation)
no. of moles A
x Molar Mass B
Product – a substance formed as a result of a chemical reaction.
Reactant – the starting materials in a chemical reaction.
Stoichiometry – the quantitative study of reactants and products in a chemical reaction.
Directions: Identify what is asked or described in each
item then write the letter of the correct answer.
1.
Sodium hydrochlorite (NaClO) is a common active ingredient of bleaching agents
such as ZONROX. It is produced by bubbling chlorine (Cl2) gas into a solution of
sodium hydroxide (NaOH) as represented by the chemical equation below:
Cl2 (g) + 2 NaOH (aq)
NaClO (aq) + NaCl (aq) + H2O (l)
What is the number of moles of Cl2 gas needed for every 2 moles of NaOH to
produce enough NaClO?
A. 1.0
B. 1.5
C. 2.0
D. 2.5
2. In order to determine the limiting reactant in a particular reaction, one must know
each of the following EXCEPT __________________________.
A. the coefficient of each reactant in a balanced equation
B. the molar mass of each reactant present
C. the mass of each product formed
D. the mass of each reactant present
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3. Which of the following is true about limiting reactant?
A. It is in excess amount.
B. It has the greater molar mass.
C. The reagent that is completely produced in a reaction.
D. The reagent that is completely used up in a reaction.
4. Consider the reaction below:
How many grams of ZnCl2 are obtained when 40g of Zn is consumed in the
reaction? (Molar Mass of Zn = 65.38 g/mol; Molar Mass of ZnCl2 = 136.28 g/mol)
A. 40 g
B. 65 g
C. 83 g
D. 135 g
5. Urea [(NH2)2CO] is prepared by reacting ammonia with carbon dioxide:
2NH3(g) + CO2(g) → (NH2)2CO(aq) + H2O(l)
In one process, 637.2 g of NH3 are treated with 1142 g of CO2. What is the
limiting reagent and what is the mass of(NH2)2CO formed?
A.
B.
C.
D.
NH3; 26 g (NH2)2CO
NH3; 1124 g (NH2)2CO
CO2; 26 g (NH2)2CO
CO2; 1124 g (NH2)2CO
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Physical Science Teaching Guide for Senior High School
Project EASE Chemistry
Physical Science. Helen E. Caintic 2016
Physical Science Dennis G. Caballes et.al. 2016
https://lrmds.deped.gov.ph/
https://www.academia.edu/Chemistry_4th_Edition_By_Chang
www.khanacademy.org
www.quipper.com
www.quizlet.com
https://www.youtube.com/watch?v=RTUFPjliMCU
https://www.youtube.com/watch?v=ymCZ2ShhBAw
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